Solve the exponential equation for $x$. 32 x 5 = ( 1 16 ) 4 x − 3 32\^{\frac x5}=\left(\dfrac{1}{16}\right)\^{ 4x-3} $x=$
Solution: The strategy We want to rewrite both of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. Matching the bases Let's rewrite 32 x 5 32\^{ \frac x5} and ( 1 16 ) 4 x − 3 \left(\dfrac{1}{16}\right)\^{ 4x-3} so their common base is $2$. 32 x 5 ( 2 5 ) x 5 2 x = ( 1 16 ) 4 x − 3 = ( 2 − 4 ) 4 x − 3 = 2 − 16 x + 12 ( 32 = 2 5 and 1 16 = 2 − 4 ) ( ( a n ) m = a n ⋅ m ) \begin{aligned}32\^{\frac x5}&=\left(\dfrac{1}{16}\right)\^{ 4x-3}\\\\ (2^5)\^{ \frac x5} &=(2^{-4})\^{ 4x-3}&&&&(32=2^5 \text{ and } \dfrac{1}{16} = 2^{-4}) \\\\ 2\^{ x} &= 2\^{-16x+12} &&&&((a^n)^m=a^{n\cdot m})\end{aligned} Solving the linear equation We obtain the following equation. 2 x = 2 − 16 x + 12 2\^{ x} = 2\^{ -16x+12} Now we can equate the exponents and solve for $x$. $\begin{aligned} x&=-16x+12\\\\ x &= \dfrac{12}{17}\end{aligned}$ The answer The answer is $x=\dfrac{12}{17}$. You can check this answer by substituting $\it{x=\dfrac{12}{17}}$ in the original equation and evaluating both sides.